Question: A particle moving in the $xy$ -plane has velocity vector given by $v(t)=\left(2t^3,3t^2\right)$ for time $t\geq 0$. At $t=1$, the particle is at the point $(3,4)$. What is the particle's position at $t=3$ ? $($
To find the particle's position at $t=3$, we need to find its horizontal displacement $\Delta x$ and its vertical displacement $\Delta y$, and add those to its initial position $(3,4)$ : $\text{Position at }t=3\text{: }(3+\Delta x,4+\Delta y)$ The particle's horizontal displacement can be found by taking the definite integral of the horizontal component of $v(t)$ between $t=1$ and $t=3$ : $\Delta x=\int_1^3 2t^3\,dt=40$ The particle's vertical displacement can be found by taking the definite integral of the vertical component of $v(t)$ between $t=1$ and $t=3$ : $\Delta y=\int_1^3 3t^2\,dt=26$ Now we can find the particle's position: $\begin{aligned} &\phantom{=}(3+\Delta x,4+\Delta y) \\\\ &=(3+40,4+26) \\\\ &=(43,30) \end{aligned}$ In conclusion, particle's position at $t=3$ is $(43,30)$.